The best way to approximate $15^{1/4}$

Question

I'm trying to approximate $\sqrt[4]{15}$ using the binomial theorem, however the simplest way I can think to do this is to separate this into ${(10\times1.5)}^{1/4}={10}^{1/4}\times{1.5}^{1/4}$.

Obviously we can approximate ${1.5}^{1/4}$ using newtowns version of the binomial method: $(1+.5)^{1/4}\approx1.10668$.

Then I was thinking I should use the Newton-Raphson method to obtain an approximation for $10^{1/4}$.This does work out (I've checked it on my calculator) but is there a way to approximate $15^{1/4}$ using just the binomial theorem?

Answer 1

As mentioned in the comments (though I did not see them while typing this): $$15^{1/4}=16^{1/4}\cdot\left(\frac{15}{16}\right)^{1/4}=2\cdot\left(1-\frac1{16}\right)^{1/4}$$ Thus the binomial series can be applied to $\left(1-\frac1{16}\right)^{1/4}$, the result to be doubled afterwards. $\frac1{16}$ is small, so this will converge quickly.

Answer 2

You can try to use this approximation $$f(x+h)\doteq f(x)+hf'(x)$$ if $h$ is suficently small. Take $f(x)=\sqrt[4]{x}$, $x=16$ and $h=-1$. So we get $$ \sqrt[4]{x+h} \doteq \sqrt[4]{x} +{h\over 4\sqrt[4]{x^3}}$$

So $$ \sqrt[4]{15} \doteq \sqrt[4]{16} -{1\over 4\sqrt[4]{16}^3}= 2-{1\over 32}$$

Answer 3

For $x$ close to the origin we have $(1-x)^{1/4} \approx 1-\frac{x}{4}-\frac{3x^2}{32}$ and an even better approximation is $(1-x)^{1/4} \approx 1-\frac{x}{4}-\frac{9x^2}{96-56x}$. By evaluating at $x=\frac{1}{16}$ and multiplying by $2$ we get the approximation $\sqrt[4]{15}\approx \color{red}{\frac{23301}{11840}}$ which is correct up to six figures.